The book I'm working from provides the answer $a=-1, b=1, c=1, d=6$, but I'm having a hard time getting there.

This is in order to reduce $q(x)=x^4+6x^2-5x+6$ to a pair of irreducible quadratics $(x^2+ax+b)(x^2+cx+d)$. We expand, gather like terms, and equate the coefficients of the resulting expression with those of the original, yielding the above system of equations.

With some substitutions, I managed to get
$$b^6-6b^5-6b^4+47b^3-36b^2-216b+216=0$$
which I reduced to
$$(b-6)(b-1)(b^4+b^3-5b^2+6b+36)=0$$
Since the remaining factor $(b^4+b^3-5b^2+6b+36)$ has no real zeros, it also reduces to a pair of irreducible quadratics. I started trying to find them with the same strategy as above, producing yet another system of equations, but I suspect I'm headed down a recursive rabbit-hole.

2 Answers
2

You were so close! You noted that the quartic factor has no real zeros, so you can conclude that $b=1$ or $b=6$, whence $d=6$ or $d=1$ (respectively) by the last equation of your system. Since $c=-a$ by the first equation of your system, then the third equation of your system becomes $a(d-b)=-5$, which gives you $a$, and thus $c=-a$.

Both $(a,b,c,d)=(-1,1,1,6)$ and $(a,b,c,d)=(1,6,-1,1)$ work as solutions.

Do you know Gauss's lemma? A polynomial with integer coefficients will factorise over the rational numbers if and only if it will factorise over the integers. That is, your values for $a, b, c, d$ must be integers.

($bd = 6$, so for an easier solution you can just guess $b = \pm 1, \pm 2, \pm 3, \pm 6$ and see what works. Of course, by symmetry, if $b = -1$ then $d = -6$ and so on, so you only really need to try half of them.)